Finite additivity of a set function ($F(A\cup B)=F(A)+F(B)(A\cap B=\varnothing )$)

A real-valued set function $F$ on an algebra of sets is finitely additive whenfor disjoint members of the algebra. Taking both sets empty gives $F(\varnothing )=0$, and induction extends the identity to every finite disjoint union.

On the full power set of a finite set $V$, every subset is a disjoint union of singletons, soThus arbitrary signed weights on the points describe exactly these finitely additive set functions. Summing the pointwise indicator function identity for the inclusion-exclusion principle against those weights gives the corresponding identity for $F$.

The full-power-set qualification matters: on the smaller algebra $\{\varnothing ,\{a,b\}\}$, the value $F(\{a,b\})$ does not specify $F(\{a\})$ or $F(\{b\})$, because the singletons are outside the domain. One must not apply the singleton formula to undefined values.